# 2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1

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## Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1

### 2nd PUC Maths Linear Programming NCERT Text Book Questions and Answers Ex 12.1

Solve the following Linear Programming Problems graphically:

Question 1.
Maximize Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥0.
x + y = 4

 X 2 3 1 y 2 1 3

put x = 0 y = 0
since 0 + 0 ≤ 0, equation is true.
∴ This will be shaded towards 0.
At A (4,0), Z = 3 x 4 = 12
At B (0,4), Z = 4 x 4= 16
∴  Z is maximised at B (0,4) = 16  Question 2.
Minimize Z = -3x + 4y
subject to x + 2y ≤ 8; 3x + 2y ≤ 12; x ≥ 0, y ≥ 0.
x + 2y = 8

 X 2 4 0 Y 3 2 4

0 + 0 < 8
true towards 0

3x +  2y = 12

 X 2 4 0 y 9 0 6

0 + 0 <12
ABCD is the solution region
x + 2y = 8
Put x = 0 ⇒ A(-0,4)
Put y = 0 ⇒ B(8,0)
At A (0,4) ⇒ Z= 16
At B (2, 3) ⇒ Z =-3 x 2+ 4 x 3 = 6
At C (4, 0) ⇒ Z = -12
At D (0,0) ⇒ z = 0
Hence Z is minimum at C (4, 0) = -12  Question 3.
Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15,5x +2y ≤ 10, x ≥ 0,y ≥ 0
3x + 5y ≤ 15

 X 0 5 y 3 0

0 + 0 <15
true, toward 0 5x +2y = 10

 X 2 0 y 0 5

ABCD is the solution region
At A(0,3) ,Z = 3 x 3 = 9

Question 4.
Minimum Z = 3x + 5y
such that x +3y ≤ 3, x+y ≥ 2, x, y  ≥0.
x + 3y = 3

 X 3 0 y 0 1

0 + 0,3 false x + y = 2

 X 0 2 1 y 2 0 1

0 + 0 ,2 false
Away from 0
corner points of feasible region are
A (0,2); Z = 5 x 2 = 10
B (1.5, 0.5); Z = 7
C (3,0); Z = 3 x 3 = 9
∴ Z is minimum at B (1.5,0.5) = 7. Question 5.
Maximise Z = 3x +2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
x + 2y = 10

 X 0 6 4 y 5 4 3

0 + 0 < 10
True 3y + y = 15

 X 5 4 3 y 0 3 6

0 + 0 < 15
True

ABCO is the solution region
At A (0,5) Z = 2 x 5 = 10
At B (4, 3) Z = 3 x 4+ 2 x 3 = 18
At C (5,0) Z = 3 x 5 = 15
∴ Z is maximized at B (4, 3) = 18.

Question 6.
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0
2x + y = 3

 X 1 0 y 1 3

0 + 0, 0 false x + 2y = 6

 X 6 0 2 y 0 3 2

0 + 0 ,6 false
An and above O is the solution region
At A(0,3) Z = 2 x 3 = 6
At B(6,0) Z = 6
Z is maximum at all points on line AB = 6
Here minimum of Z occurs at 2 points A & B. Show that the minimum of Z occurs at more than two points.

Question 7.
Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0,x,y ≥ 0.
x + 2y = 120

 X 60 0 20 Y 30 60 50

0 + 0 ,120
True

x + y = 60

 X 30 20 40 Y 30 40 20

0 + 0 >0
false x – 2y =0

 x 20 0 40 y 10 0 20

40 – 20 > 0
ABCD is the solution region
At A (60, 30) Z = 5 x 60 + 10 x 30 = 600
At B (40, 20) Z = 5 x 40+10 x 20 = 400
At C (60,0) Z = 5 x 60 = 300
At D (120,0) Z = 5 x 120 = 600
Z is minimized at C (60,0) = 300
Z is maximized at 2 points A (60,30), D (120,0)
∴ maximum value is 600 at all
points on the line joining A (60,30) and D (120,0) Question 8.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
x + 2y = 100

 x 100 50 20 y 0 20 40

0 + 0>100
0 – 0 < 0 False

2x – y = 0

 x 100 20 10 y 200 40 20

0 – 0 < 0
False

2x +y = 200

 x 100 50 60 y 0 100 80

0 + 0 <200
True ABCD is the solution                              .
At A (0,200),          Z = 2 x 200 = 400
At B (50,100), Z = 50 + 2 x 100 = 250
At C (20,40), Z = 20 + 2x 40= 100
At D (0,50), Z = 2 x 50 =100
Z is maximised at A (0,200) = 400
Z is minimised at 2 points C (20,40) & D (0, 50) = 100 Question 9.
Maximise Z = -x + 2y, subject to the constraints:
x ≥ 3, x + y ≥ 5 , x + 2y ≥ 6, y ≥ 0
x + y = 5

 x 1 3 2 y 4 2 3

0 + 0, 5
False

0,3
False x + 2y = 6

 x 0 4 2 y 3 1 2

0 + 0
False

ABC is the solution region
At A (3,2)       Z =-3+2 x 2=1
At B (4,1)       Z = -4 + 2 x 1 = -2
At C (6, 0)      Z = -6 = -6
Z is maximised at A (3,2) = 1 and
minimised at C (6,0) = -6 Question 10.
Maximise Z = x + y,
subject to x – y ≤ -1, -x +y ≤ 0, x, y ≥ 0.
x – y = -1

 x 1 0 2 y 2 1 3

0 – 0 ,-1
False

x + 2y = 6

 x 0 4 2 y 3 1 2

0 + 0 , 0 False,
Since there is no common shaded region, there is no feasible solution region. Hence maximum of Z does not exist. 